Are the trigonometric limits such as
$\lim_{x\to 0}\frac{sin(x)}{x} = 1$ valid only for all positive $x$ ?
For example , my book says that
$\lim_{x\to 0}\frac{sin(-1)}{-1} = sin1$ instead of just $1$.
So what is the condition for these limits to be valid ? Thank you for your help
On
We have $\displaystyle \lim_{x\to 0^+} \frac{\sin x}{x}=1$ and $\displaystyle \lim_{x\to 0^-} \frac{\sin x}{x}=1$. $x$ can be negative.
$\displaystyle \frac{\sin(-x)}{-x}= \frac{-\sin(x)}{-x}=\frac{\sin(x)}{x}$ and $\displaystyle \lim_{x\to 0} \frac{\sin (-x)}{-x}=1$.
$\displaystyle \frac{\sin(-1)}{-1}$ is a constant and it is always equal to $\sin 1$. So we obviously have $\displaystyle \lim_{x\to 0} \frac{\sin(-1)}{-1}=\sin 1$.
On
Note that $\sin x$ and $x$ are both odd functions thus $\frac{\sin x}{x}=\frac{\sin -x}{-x}$ and the limit for $x\to 0^-$ and for $x\to 0^+$ is the same.
On
For $$\lim_{x\to 0}\frac{sin(-1)}{-1} =\lim_{x\to 0}\frac{-sin(1)}{-1} = sin(1)$$
You are taking limit of a constant which is the same as the constant itself.
For $$\lim_{x\to 0}\frac{\sin(x)}{x} = 1$$
You are taking the limit of a function and you need some mathematics to prove it.
The limit exist because you get the same result for positive or negative x.
No, in fact: $x$ isn't limited to positive values in the limit above... It can be positive and negative when tending to $0$.
But there's no variable $x$ here! This is similar to writing: $$\lim_{x \to 0}\frac{-5}{4} = -\frac{5}{4}$$ because $\tfrac{-5}{4}$ is simply a constant, not depending on $x$.
The same holds for your example, the only difference being that this constant involves a sine-function. Note that the sine function is an odd function which means that $\sin(-a)=-\sin a$ for any value $a$, so also: $$\lim_{x\to 0}\frac{\sin(-1)}{-1} = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1}=\sin 1$$ Although it may look similar, this has nothing to do with the (standard) limit: $$\lim_{x\to 0}\frac{\sin(x)}{x} = 1$$